Implicit multivariable derivative of a spherical cap

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Homework Statement



Consider a spherical cap, for which the surface area and volume is

[tex]A(a,h) = \pi(a^2 +h^2)[/tex]
[tex]V(a,h) = \frac{\pi h}{6}(3a^2 +h^2)[/tex]

What would the aspect ratio [itex]dA/dV[/itex] be?

The Attempt at a Solution



Clearly we would have

[tex]dA = 2\pi a da + 2\pi h dh[/tex]
[tex]dV = \pi ha da + \frac{\pi}{2}(a^2+h^2) dh [/tex]

but what would the chain rule look like? Surely, it should have more terms than just

[tex]\frac{dA}{dV } = \frac{\partial A}{\partial a} \frac{\partial a}{\partial V} + \frac{\partial A}{\partial h}\frac{\partial h}{\partial V}[/tex]

, right?
 

Answers and Replies

  • #2
HallsofIvy
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Why do you think it should have more terms? a and h are the only variables.
 
  • #3
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If [itex]dV, dA[/itex] were just functions of one variable, it would be straightforward. But here there are two variables, and perhaps I would have anticipated a chain rule of the form

[tex]\frac{dA}{dV} = \frac{\partial A}{\partial a}\left( \frac{\partial a}{\partial V} + \frac{\partial h}{\partial V} \right) + \frac{\partial V}{\partial a}\left( \frac{\partial a}{\partial A} + \frac{\partial h}{\partial A} \right) [/tex]

but I'm just guessing.
 
  • #4
pasmith
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On dimensional grounds, I would expect [tex]
A = V^{2/3} f\left(\frac ah\right)[/tex] for some function [itex]f[/itex] which can be determined from the formulae for [itex]A[/itex] and [itex]V[/itex]. Thus [itex]A[/itex] is a function of both [itex]V[/itex] and [itex]a/h[/itex].
 
  • #5
HallsofIvy
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If [itex]dV, dA[/itex] were just functions of one variable, it would be straightforward. But here there are two variables, and perhaps I would have anticipated a chain rule of the form

[tex]\frac{dA}{dV} = \frac{\partial A}{\partial a}\left( \frac{\partial a}{\partial V} + \frac{\partial h}{\partial V} \right) + \frac{\partial V}{\partial a}\left( \frac{\partial a}{\partial A} + \frac{\partial h}{\partial A} \right) [/tex]

but I'm just guessing.
The second part has V "in the numerator" that can't be right! "On dimensional grounds", as pasmith puts it, V has units of "length cubed", A has units of "length squared, and h and a have units of "length". [itex]\partial V/\partial a[/itex] has units of "length squared", and [itex]\partial a/\partial A[/itex] and [itex]\partial h/\partial A[/itex] have units of "1 over length" so the second term has units of length. But [itex]dA/dV[/itex] has units of "1 over length".
 
  • #6
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Thanks for the advice HallsofIvy and pasmith!

Pasmith, yes, you were right, I was able to evaluate the derivative by expressing the functions in terms of [itex]f(h/a)[/itex]. Thanks for the insight!
 
  • #7
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Dear Pasmith, I have been looking at this problem again and in fact I don't think it is possible to find the derivative [itex]dA/dV[/itex] like you mentioned.

Let [itex]x\equiv a/h.[/itex]
[itex]A^3 = V^2 f(a/h) = V^2 f(x)[/itex]
[itex]3A^2 dA = 2V dV f(x) + V^2 \frac{df}{dx} dx[/itex]

The complication comes in when you are trying to solve for [itex]dA/dV[/itex] and you get

[itex]\frac{dA}{dV} = ... + \frac{V^2}{3A^2} \frac{df}{dx} \frac{dx}{dV}.[/itex]

But note that [itex]V=\frac{\pi h}{6}(3a^2+h^2)[/itex]. Since the powers of [itex]h, a[/itex] are different, there is no way to evaluate the derivative [itex]dx/dV[/itex] without leaving behind some residual terms of [itex]a, h[/itex]. The goal in the first place was to evaluate the derivative solely in [itex]x\equiv a/h[/itex]. Therefore we can't parameterise the equation like you said here.

Can you help, Pasmith, or anybody? This problem has been bothering me for a while.
 
  • #8
PeroK
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A simpler approach is to use the relationship between ##a## and ##h## to express ##A## and ##V## in terms of ##h## and the radius of the sphere. That reduces it to a single variable problem.
 
  • #9
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A simpler approach is to use the relationship between ##a## and ##h## to express ##A## and ##V## in terms of ##h## and the radius of the sphere. That reduces it to a single variable problem.
[itex]a , h[/itex] are also related by a contact angle [itex]\theta[/itex]. But you can't express a spherical cap uniquely in just that one variable.

I guess I don't really understand what you meant. From what I understand a spherical cap is uniquely described by at least two parameters - one possible combination is the the base width and height [itex]a,h[/itex], and another combination is the radius of curvature and angle [itex]r,\theta[/itex].
 
  • #10
PeroK
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[itex]a , h[/itex] are also related by a contact angle [itex]\theta[/itex]. But you can't express a spherical cap uniquely in just that one variable.
I guess from what you said that this is not a spherical cap on a given sphere?
 
  • #11
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I guess from what you said that this is not a spherical cap on a given sphere?
Yes, that's right, the cap is not on a given sphere. There are two parameters.
 
  • #12
PeroK
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Yes, that's right, the cap is not on a given sphere. There are two parameters.
So, are you not trying to calculate ##\frac{\partial A}{\partial V}## ?
 
  • #13
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So, are you not trying to calculate ##\frac{\partial A}{\partial V}## ?
That is what I am calculating. As we know, for a sphere, [itex]V(r), A(r)[/itex] and it's easy to calculate [itex]\frac{dA}{dV} = \frac{dA}{dr} \frac{dr}{dV}[/itex]. But what happens when these two quantities are functions of more than one variable, ie, [itex]V(a,h), A(a,h)[/itex]? Then how would we calculate [itex]\partial A/\partial V[/itex]?
 
  • #14
PeroK
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That is what I am calculating. As we know, for a sphere, [itex]V(r), A(r)[/itex] and it's easy to calculate [itex]\frac{dA}{dV} = \frac{dA}{dr} \frac{dr}{dV}[/itex]. But what happens when these two quantities are functions of more than one variable, ie, [itex]V(a,h), A(a,h)[/itex]? Then how would we calculate [itex]\partial A/\partial V[/itex]?
I think your problem is that you also need to specify which variable is constant for your partial derivative. For example, you could express A as a function of V and h, say:

##A = f(V, h)##

And then you could calculate [itex]\partial A/\partial V = \partial f/\partial V [/itex] where h is held constant.

But, you could also find:

##A = g(V, a)##

And then you could calculate [itex]\partial A/\partial V= \partial g/\partial V [/itex] where a is held constant. And these would be different functions with different numerical values.

Or, you could hold ##a/h## constant etc.

That's why I assumed it was a fixed sphere, when you have V as a function of A with no other independent variables and can find ##\frac{dA}{dV}##
 
  • #15
Ray Vickson
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Homework Statement



Consider a spherical cap, for which the surface area and volume is

[tex]A(a,h) = \pi(a^2 +h^2)[/tex]
[tex]V(a,h) = \frac{\pi h}{6}(3a^2 +h^2)[/tex]

What would the aspect ratio [itex]dA/dV[/itex] be?

The Attempt at a Solution



Clearly we would have

[tex]dA = 2\pi a da + 2\pi h dh[/tex]
[tex]dV = \pi ha da + \frac{\pi}{2}(a^2+h^2) dh [/tex]

but what would the chain rule look like? Surely, it should have more terms than just

[tex]\frac{dA}{dV } = \frac{\partial A}{\partial a} \frac{\partial a}{\partial V} + \frac{\partial A}{\partial h}\frac{\partial h}{\partial V}[/tex]

, right?
Your question is not well-defined (as many others have also indicated). However, it is not very difficult to extract answers to more pointed questions. We have
[tex] \Delta A = 2 \pi a \, \Delta a + 2 \pi h \, \Delta h, \\
\Delta V = \pi h a \, \Delta a + \frac{\pi}{2}(a^2+h^2) \, \Delta h \\
\text{so}\\
\frac{\Delta A}{\Delta V} =
\frac{2 \pi a \, \Delta a + 2 \pi h \, \Delta h}{\pi h a \, \Delta a + (\pi /2)(a^2+h^2) \, \Delta h}
[/tex]
This implies the following.
(1) If ##a## is a function of ##h## we get
[tex] \frac{dA}{dV} = \frac{2 \pi a (da/dh) + 2 \pi h}{\pi h a (da/dh) + (\pi/2) (a^2 + h^2)}[/tex]
(2) If ##h## is a function of ##a## we get
[tex] \frac{dA}{dV} = \frac{2 \pi a+ 2 \pi h (dh/da)}{\pi h a + (\pi/2)(a^2+h^2) (dh/da)} [/tex]
(3) If ##h## is a constant, the derivative ##(dA/dV)_h## --which is Thermodynamics-style notation, indicating that ##h## is held constant--has the form
[tex] \left( \frac{d A}{d V} \right)_{h} = \frac{ 2 \pi a}{\pi h a}.[/tex]
(4) If ##a## is a constant, the derivative ##(dA/dV)_a## is
[tex] \left( \frac{dA}{dV}\right)_a = \frac{2 \pi h}{(\pi/2)(a^2+h^2)}[/tex]

You get many different answers, depending on what you mean by the question to begin with.
 

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